Question

If $\log 2$, $\log (2^x-1)$ and $\log (2^x+3)$ (all to the base $10$) are three consecutive terms of an Arithmetic Progression, then the value of $x$ is equal to :

• A. $0$
• B. $1$
• C. ${\log }_{2}5$
• D. ${\log }_{10}2$

Answer 1

Answer: (C)

${\log }_{2}5$

$\log 2,\log (2^x-1),\log (2^x+3)$ are in A.P.

$\Rightarrow 2\left[\log \right(2^x-1\left)\right]=\log 2+\log (2^x+3)$

$=\log [2\times (2^x+3\left)\right]$

$\Rightarrow \log (2^x-1{)}^2=\log (2^{x+1}+6)$

$\Rightarrow (2^x-1{)}^2=2^{x+1}+6=2^x.2+6$

Let $2^x=y$

$\therefore (y-1{)}^2=2y+6$

$\Rightarrow y^2-2y+1=2y+6$

$\Rightarrow y^2-4y-5=0$

$\Rightarrow (y-5)(y+1)=0$

$\Rightarrow y=5,-1$

When $y=5$

$\Rightarrow 2^x=5\Rightarrow x\log 2=\log 5$

$\Rightarrow x=\frac{\log 5}{\log 2}$

$\Rightarrow x={\log }_{2}5$